Java8实战之Stream的延迟计算

    public static List<List<Integer>> findAllSubList(List<Integer> list) {
        if (list.size()==0) {
            List<List<Integer>> res = new ArrayList<>();
            res.add(Collections.emptyList());
            return res;
        }
        Integer first = list.get(0);
        List<Integer> subList = list.subList(1, list.size());
        List<List<Integer>> allSubList = findAllSubList(subList);
        List<List<Integer>> allSubList2 = insertAll(first, allSubList);
        return concat(allSubList, allSubList2);
    }
    
    private static List<List<Integer>> concat(List<List<Integer>> allSubList, List<List<Integer>> allSubList2) {
        List<List<Integer>> res = new ArrayList<>(allSubList);
        res.addAll(allSubList2);
        return res;
    }

    private static List<List<Integer>> insertAll(Integer item, List<List<Integer>> allSubList) {
        List<List<Integer>> res = new ArrayList<>();
        for (List<Integer> a : allSubList) {
            List<Integer> oneList = new ArrayList<>(a);
            oneList.add(item);
            res.add(oneList);
        }
        return res;
    }

    /**
     * 使用迭代计算阶乘
     * r 和 i 在每轮迭代中都会更新
     * @param n
     * @return
     */
    public static int factorialIterator(int n) {
        int r = 1;
        for (int i=1; i<=n; i++) {
            r *= i;
        }
        return r;
    }

    /**
     * 使用递归 计算阶乘
     *  比迭代都效率差：因为每次递归都需要创建栈桢
     * @param n
     * @return
     */
    public static int factorialRecursive(int n) {
        return n==1? 1 : n * factorialIterator(n-1);
    }

    /**
     * 使用Stream 计算阶乘
     * @param n
     * @return
     */
    public static int factorialStream(int n) {
        return IntStream.rangeClosed(1, n).reduce(1, (x, y)->x*y);
    }

    /**
     * 尾-递 迭代
     * @param n
     * @return
     */
    public static int factorialTailIterator(int n) {
        return factorialTailHelp(1, n);
    }

    /**
     * 尾-递 迭代递帮助类
     * @param acc
     * @param n
     * @return
     */
    private static int factorialTailHelp(int acc, int n) {
        return n==1?acc:factorialTailHelp(acc*n, n-1);
    }

/**
 * @Date 2021/9/5
 * @Author lifei
 */
public interface MyList<T> {

    T head();
    MyList<T> tail();

    MyList<T> filter(Predicate<T> p);

    default boolean isEmpty() {
        return true;
    }
}

public class LazyList<T> implements MyList<T> {
    final T head;
    final Supplier<MyList<T>> tail;

    public LazyList(T head, Supplier<MyList<T>> tail) {
        this.head = head;
        this.tail = tail;
    }
    @Override
    public T head() {
        return head;
    }

    @Override
    public MyList<T> tail() {
        return tail.get();
    }

    @Override
    public MyList<T> filter(Predicate<T> p) {
        return isEmpty()?this:p.test(head())? new LazyList<>(head, ()->tail().filter(p)):tail().filter(p);
    }

    @Override
    public boolean isEmpty() {
        return false;
    }
}

    /**
     * 创建一个无限延迟的列表
     * @param n
     * @return
     */
    public static LazyList<Integer> from(int n) {
        return new LazyList<>(n, ()->from(n+1));
    }

    /**
     * 创建一个无限循环的 质数列表
     * @param numbers
     * @return
     */
    public static MyList<Integer> primes(MyList<Integer> numbers) {
        return new LazyList<>(numbers.head(), ()->primes(numbers.tail().filter(n->n%numbers.head()!=0)));
    }

    public static void main(String[] args) {
        LazyList<Integer> numbers = from(2);
        Integer res2 = numbers.head();
        Integer res3 = numbers.tail().head();
        Integer res4 = numbers.tail().tail().head();
        System.out.println(res2);
        System.out.println(res3);
        System.out.println(res4);
        System.out.println("创建一个无限延迟的质数列表");
        MyList<Integer> primes = primes(numbers);
        for (int i=0; i<30; i++) {
            if (!primes.isEmpty()){
                System.out.print(primes.head() + ", ");
                primes = primes.tail();
            }
        }
        System.out.println();

    }